# Expected value of X*Y

I'm stuck on the following probability problem and would welcome any help:

Consider three random uniform variables on [0,1]. Let X be the minimum of these three variables and Y be the maximum. What is the expected value of X*Y?

Am I right in assuming I need to find the joint distribution function of X and Y? I have found the separate distribution functions but I'm having trouble with the joint one since X and Y are not independent.

• And... what did you try to find the joint distribution of $(X,Y)$? – Did Jan 22 '19 at 17:15
• The distribution function for X is: $P(X<x) = 1-(1-x)^3$. Likewise, $P(Y<y) = y^3$. So for the joint function, $P(X<x, Y<y)$ I tried separating the cases $x>y$ and $x<y$. Indeed, if $x>y$ then, $P(X<x,Y<y) = P(Y<y) = y^3$ since in this case $X<Y<y<x$. For the case $x<y$ I'm kind of lost, but it is perhaps possible that here X and Y are independent, in which case we have $P(X<x,Y<y) = y^3*(1-(1-x)^3)$. If you can help find the joint function I'd be very grateful, although I now realise that passing through joint functions is not the easiest solution. – nicdel Jan 23 '19 at 0:11
• $X$ and $Y$ are certainly not independent, because both are concentrated on $[0,1]$ but $X \le Y$ always. – Robert Israel Jan 23 '19 at 1:21

It's simpler to deal with the three uniform random variables $$U_1, U_2, U_3$$. Given $$U_1 < U_2 < U_3$$, $$XY = U_1 U_3$$. So \eqalign{ \mathbb E [XY | U_1 < U_2 < U_3] &= \mathbb E[U_1 U_3 | U_1 < U_2 < U_3]\cr &= 6 \int_0^1 du_3 \int_0^{u_3} du_2 \int_0^{u_2} du_1 \; u_1 u_3} By symmetry, the result is the same for all the other orderings of $$U_1, U_2, U_3$$, so this is also $$\mathbb E[XY]$$.